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Mathematics 20 Online
OpenStudy (colcaps):

Given: f(x) = 2x2 – 3x + 4 g(x) = x + 3 h(x) = (2x-1)/(x+2) k(x) = (x-1)/(x+1) Find: ( f o g o h o k)( 1 )

OpenStudy (anonymous):

It will take time..wait:D

OpenStudy (colcaps):

whats "o"?

OpenStudy (anonymous):

composition

OpenStudy (anonymous):

fog means f ( g(x) )

OpenStudy (anonymous):

its times lol..hehe..times = composition :)

OpenStudy (colcaps):

what if its a solid circle?

OpenStudy (anonymous):

@pengembara_bumi1 times??

OpenStudy (anonymous):

multiply?

OpenStudy (colcaps):

composition is multiplication?

OpenStudy (anonymous):

no @Colcaps

OpenStudy (anonymous):

start your calculation by insert the function of k(1) first...then continue with h(k)..then g(h)..finally f(g)...hehe

OpenStudy (anonymous):

@Krishnadas : no...i mean...u substitute 1 into k(x)..then follow the rule..hehe

OpenStudy (anonymous):

@pengembara_bumi1 can you please show him how its solved?I am going

OpenStudy (anonymous):

@Colcaps : dont missunderstood..hehe..

OpenStudy (colcaps):

Wait im confused, what will "(1)" do in there?

OpenStudy (anonymous):

@Krishnadas : orite :)

OpenStudy (anonymous):

the answer is 9

OpenStudy (colcaps):

Wait, i dont want the answer, i want to know how its solved

OpenStudy (anonymous):

first plug 1 in the place of x in k(x). FInd the value and then plug that value in h(x) and so on till you come to f(x) it will give you the answer.

OpenStudy (anonymous):

k(1) = 1-1/1+1 = 0

OpenStudy (anonymous):

now find the value of h(x) when x = 0

OpenStudy (colcaps):

So k(1)=x-1/x+1?

OpenStudy (anonymous):

the value that that you get for h(0) is the value you plug in the place of x in g(x)

OpenStudy (anonymous):

k(x) = x-1/x+1, when you put 1 in the place of x it becomes k(1)

OpenStudy (colcaps):

So what will happen if its changed to k(1)?

OpenStudy (anonymous):

you have to find the value of (fogohok) (1)so the easiest way is to first find the value of k(1) then use this value and put it in the place of x in the function h(x)

OpenStudy (anonymous):

for example, when you put x = 1 in k(x) it becomes 0; k(1) = 0

OpenStudy (anonymous):

f(x) = 2x2 – 3x + 4 g(x) = x + 3 h(x) = (2x-1)/(x+2) k(x) = (x-1)/(x+1) =f.g.h ( (1-1)/(1+1) ) =f.g.(2(0)-1)/(0+2) =f.g(-1/2) =f(-1/2)+3 =f(5/2) =2(5/2)^2 - 3(5/2) + 4 =9 :)

OpenStudy (anonymous):

@akitav : am i right ? :O

OpenStudy (colcaps):

wow i really suck in combning

OpenStudy (anonymous):

now put x = 0 in the function h(x)

OpenStudy (anonymous):

@pengembara_bumi1 nice work :)

OpenStudy (colcaps):

what happened to all "x's"?

OpenStudy (anonymous):

when you need to find the value of a function at x = 1 the x would be replaced by 1. This is what we did for k(1)

OpenStudy (anonymous):

@akitav : thanks.. :)

OpenStudy (colcaps):

all "x" will be replaced to 1?

OpenStudy (anonymous):

yes dear..replace it..once u replace x with number,then no more x exist :)

OpenStudy (colcaps):

x will always be replaced to 1?

OpenStudy (colcaps):

@pengembara_bumi1

OpenStudy (anonymous):

yes..totally :)

OpenStudy (colcaps):

what is f(2) will i replace all "x" to 2?

OpenStudy (anonymous):

ofcourse.. :)

OpenStudy (colcaps):

all right, thanks

OpenStudy (colcaps):

@pengembara_bumi1 oh i forgot, this "o" is composition right? what if its a solid circle? like this 3[(g . h . k)(2)]?

OpenStudy (anonymous):

are you clear ? ok :)

OpenStudy (colcaps):

@pengembara_bumi1 oh i forgot, this "o" is composition right? what if its a solid circle? like this 3[(g . h . k)(2)]?

OpenStudy (anonymous):

ouh,also same dear :)

OpenStudy (colcaps):

alright thanks

OpenStudy (anonymous):

welcome :)

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